Question

Calculate the molarity of 0.400 mol of Na, S in 1.30 L of solution. molarity: Calculate the molarity of 23.9 g of MgS in 843 mL of solution. molarity:

Answer 1

Answer: The molarity of sodium sulfide and magnesium sulfide solution is 0.308M and 0.503 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$    .....(1)

Or,

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$     ......(2)

• For sodium sulfide:

Moles of sodium sulfide = 0.400 moles

Volume of solution = 1.30 L

Putting values in equation 1, we get:

$\text{Molarity of }Na_2S=\frac{0.400mol}{1.30L}=0.308M$

Hence, the molarity of sodium sulfide solution is 0.308 M

• For magnesium sulfide:

Mass of MgS = 23.9 g

Molar mass of MgS = 56.4 g/mol

Volume of solution = 843 mL

Putting values in equation 2, we get:

$\text{Molarity of MgS solution}=\frac{23.9\times 1000}{56.4\times 843}\\\\\text{Molarity of MgS solution}=0.503M$

Hence, the molarity of magnesium sulfide solution is 0.503 M