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3 years ago

If 31.8 grams of LiCl are dissolved in 225 grams of water, what is the concentration of the solution in percent by mass?

Chemistry

2 answers:

erastovalidia [21]3 years ago

8 0

(31.8/(225+31.8))(100)=12.38317757% LiCl

melomori [17]3 years ago

3 0

Answer: 12.4 %

Explanation:

To calculate the mass percent of solution, we use the formula:

$\text{Mass percent of solution}=\frac{\text{Mass of solute}}{\text{mass of solution}}\times 100$

Given :

mass of solute (LiCl) = 31.8 grams

mass of solvent (water) = 225 grams

Mass of solution = mass of solute + mass of solvent = 31.8 g + 225 g= 256.8 g

$\text{Mass percent of solution}=\frac{31.8}{256.8g}\times 100=12.4%$

Thus concentration of the solution in percent by mass is 12.4%

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3 years ago

Read 2 more answers

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

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3 years ago

How many electron are in 56fe 3

RSB [31]

The periodic table is arranged in a way so that with each step the number of protons in the nucleus is increased by 1. It makes it for an easy choice to designate elements with numbers - atomic numbers, because in that case atomic number shows the number of protons possessed by the nucleus. Like this:
H has 1 proton
He has 2 protons
Li has 3 protons
Be has 4 protons and so on

Each proton has a charge of +1. The other particle present in the nucleus - the neutron - has zero electrical charge and thus irrelevant when computing the charge of a nucleus. It is easy to deduce that the nucleus charge equals the number of protons (which in turn equals the atomic number). So the nucleus charges are:
for H it's+1
for He it's +2
for Li it's +3
for Be it's +4 and so on

Atom is an electroneutral particle by definition. It means it's summed charge must be 0. Since we've looked at everything within the nucleus (the protons and the neutrons) it's time we turn our gaze to the space around it, which is full of orbiting electrons. Each electron has a charge of -1. To make up for the positive charge in the nucleus you have to fill the space aroung the nucleus with negative electrons.Thanks to the elementary nature of both proton and electron charge, you simply have to take the same number of electrons as that of protons! Like this:
H has 1 proton and 1 electron
He has 2 protons and 2 electrons
Li has 3 protons and 3 electrons
Be has 4 protons and 4 electrons and so on

Fe has atomic number 26. It means that Fe has 26 protons and 26 electrons. If it's a neutral atom

You typed 3. Is it accidental? If so, then the answer is above. If not, then you could be trying to type 56Fe +3, which means an ionic iron with charge +3. Charges are formed when you have too many or too few electrons to counter-balance the prositive charge of the nucleus. Charge +3 means you're 3 electrons short to negate the nucleus positive charge.

In other words, Fe+3 has 26 protons and 23 electrons.

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3 years ago