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.

Step-by-step 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}* 1000}{\text{Molar mass of solute}* \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=(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}=(23.9* 1000)/(56.4* 843)\\\\\text{Molarity of MgS solution}=0.503M`

Hence, the molarity of magnesium sulfide solution is 0.503 M