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lara31 [8.8K]
2 years ago
7

An aqueous solution of 4.57 M H2SO4 has a density of 1.25 g/mL. Calculate the molality of this solution

Chemistry
2 answers:
Liono4ka [1.6K]2 years ago
8 0

Answer : The molality of solution is, 5.69 mole/L

Explanation :

The relation between the molarity, molality and the density of the solution is,

d=M[\frac{1}{m}+\frac{M_b}{1000}]

where,

d = density of solution  = 1.25 g/mL

m = molality of solution  = ?

M = molarity of solution  = 4.57 M

M_b =\text{molar mass of solute }(H_2SO_4) = 98 g/mole

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

1.25g/ml=4.57M[\frac{1}{m}+\frac{98g/mole}{1000}]

m=5.69mol/kg

Therefore, the molality of solution is, 5.69 mole/L

vladimir2022 [97]2 years ago
4 0

Answer:

The molality is 5.7 molal

Explanation:

Step 1: Data given

Molarity of H2SO4 = 4.57 M ( = 4.57 mol/L)

Density of the solution = 1.25 g/mL

Molar mass of H2SO4 = 98.09 g/mol

Step 2: Calculate mass of solution

Suppose we have 1L (= 1000 mL) of solution

Mass of the solution = 1.25 g/mL * 1000 mL

Mass of the solution = 1250 grams

Step 3: Calculate mass H2SO4

In 1L of a 4.57 M solution we have 4.57

Mass H2SO4 = moles H2SO4 * molar mass H2SO4

Mass H2SO4 = 4.57 moles * 98.09 g/mol

Mass H2SO4 = 448.3 grams

Step 4: Calculate mass solvent

Mass solvent = mass solution - mass H2SO4

Mass solvent = 1250 grams - 448.3 grams

Mass solvent = 801.7 grams

Step 5: Calculate molality

Molality = moles H2SO4 / mass solvent

Molality = 4.57 moles / 0.8017 kg

Molality = 5.7 molal

The molality is 5.7 molal

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Answer:

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

Explanation:

Step 1: Data given

Mass of sodium bicarbonate = 2.7 grams

Step 2: The balanced equation

HCl + NaHCO3 ⇔  NaCl + H2O + CO2

Step 3: Calculate moles NaHCO3

moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles

Step 4: Calculate moles HCl

For 1 mol NaHCO3 we need 1 mol HCl

For 0.032 moles NaHCO3 = 0.032 moles HCl

Step 5: Calculate mass HCl

Mass HCl = moles HCl * molar mass HCl

mass HCl = 0.032 * 36.46 g/mol= 1.17 grams

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Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9
EleoNora [17]

Answer : The current passing between the electrodes is, 1.056\times 10^{-2}A

Explanation :

First we have to calculate the charge of sodium ion.

q=ne

where,

q = charge of sodium ion

n = number of sodium ion = 2.68\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C

Now we have to calculate the charge of chlorine ion.

q'=ne

where,

q' = charge of chlorine ion

n = number of chlorine ion = 3.92\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C

Now we have to calculate the current passing between the electrodes.

I=\frac{q}{t}+\frac{q'}{t}

I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}

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4 0
3 years ago
| A solution containing 4.48 ppm KMnO4 exhibits
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Answer:

Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹

Explanation:

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⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l

⇒ Molarity = \frac{4.48 X10^{-3} }{158.03X 1(lit)} = 2.83 x 10⁻⁵ molar

Absorbance (A) = - log(T)     ( T = % transmittance)

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                          = 0.06

According to Lambert Beer's law

     

                 ε = \frac{A}{C X l}

      or,      ε = \frac{0.06}{2.83 X 10^{-5}X1 cm }

      or,      ε = 2120.14 cm⁻¹M⁻¹

Where

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We make a simplification because x<<< 0.0225:

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