Question

$6.023 \times 10^{22}$ molecules are present in 10 g of a substance X. The molarity of a solution containing 5 g of substance X in 2 L solution is ______$\times 10^{-3}$ M

Answer 1

25

Answer 2

Step 1: Calculate the number of moles of substance X from the given number of molecules.

We know that 1 mole of any substance contains Avogadro's number ($N_A = 6.023 \times 10^{23}$) of molecules.

Given: $6.023 \times 10^{22}$ molecules of substance X are present in 10 g.

Number of moles of X = (Given number of molecules) / (Avogadro's number)

Number of moles of X = $\frac{6.023 \times 10^{22}}{6.023 \times 10^{23}} = 10^{-1} = 0.1$ mol

Step 2: Calculate the molar mass of substance X.

We found that 0.1 moles of substance X weigh 10 g.

Molar mass (M) = Mass / Number of moles

M = $\frac{10 \text{ g}}{0.1 \text{ mol}} = 100 \text{ g/mol}$

Step 3: Calculate the number of moles of substance X present in 5 g.

The solution contains 5 g of substance X.

Number of moles of X = Mass of X / Molar mass of X

Number of moles of X = $\frac{5 \text{ g}}{100 \text{ g/mol}} = 0.05 \text{ mol}$

Step 4: Calculate the molarity of the solution.

Molarity (M) = Number of moles of solute / Volume of solution (in Liters)

Given volume of solution = 2 L

Molarity = $\frac{0.05 \text{ mol}}{2 \text{ L}} = 0.025 \text{ M}$

Step 5: Express the molarity in the required format ($\times 10^{-3}$ M).

$0.025 \text{ M} = 0.025 \times 10^3 \times 10^{-3} \text{ M} = 25 \times 10^{-3} \text{ M}$

The molarity of the solution is $25 \times 10^{-3}$ M.