How many grams of potassium chloride are required to make 250 ml of a 0.75 m kcl solution?

Chemistry

1 answer:

DedPeter [7]3 years ago

7 0

Answer:

The solution would need 13.9 g of KCl

Explanation:

0.75 m, means molal concentration

0.75 moles in 1 kg of solvent.

Let's think as an aqueous solution.

250 mL = 250 g, cause water density (1g/mL)

1000 g have 0.75 moles of solute

250 g will have (0.75 . 250)/1000 = 0.1875 moles of KCl

Let's convert that moles in mass (mol . molar mass)

0.1875 m . 74.55 g/m = 13.9 g

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3 years ago

A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V$

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V$