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3 years ago

The mole fraction of NaCl in an

Chemistry

1 answer:

posledela3 years ago

6 0

The weight/weight percent(%w/w) of NaCl in solution : 33.03%

<h3>Further explanation</h3>

Given

Mole fraction of NaCl : 0.132

Required

The weight/weight percent of NaCl

Solution

1 mol solution : 0.132 mol NaCl + 0.868 mol H₂O

mass NaCl :

$\tt 0.132\times 58.44 g/mol=7.714~g$

mass H₂O :

$\tt 0.868\times 18.016~g/mol=15.64~g$

Total mass = 7.714 + 15.64 = 23.354 g

$\tt \%NaCl=\dfrac{7.714}{23.354}\times 100\%=33.03\%$

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The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

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3 years ago

What is the name of the thermodynamic barrier that must be overcome before products are formed in a spontaneous reaction?

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Activation energy is a thermodynamic barrier that must be overcome before products are formed in a reaction. It is the minimum amount of energy needed for a reaction to occur. The energy can be in the form of kinetic or potential energy. This concept was introduced by Svante Arrhenius, which brought about the Arrhenius equation which is a formula used to determine rate of reactions.

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3 years ago

What is the molar mass of 37.96 g of gas exerting a pressure of 3.29 on the walls of a 4.60 L container at 375 K?

Phantasy [73]

The molar mass of the gas is 77.20 gm/mole.

Explanation:

The data given is:

P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams

Using the ideal Gas Law will give the number of moles of the gas. The formula is

PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole

Also number of moles is not given so applying the formula

n= mass ÷ molar mass of one mole of the gas.

n = m ÷ x ( x molar mass) ( m mass given)

Now putting the values in Ideal Gas Law equation

PV = m ÷ x RT

3.29 × 4.60 = 37.96/x × 0.08206 × 375

15.134 = 1168.1241 ÷ x

15.134x = 1168.1241

x = 1168.1241 ÷ 15.13

x = 77.20 gm/mol

If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.

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If one iron nail weighs 2.0 grams, what would be the mass of 6.022 x 10^23 nails?

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Answer:

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Explanation:

Hello!

In this case, it is possible to treat this problem by using a proportional factor which indicates one iron nail equals 2.0 grams:

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