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NVAEasyJEE 2024Concentration Terms (Molarity, Molality, etc.)

JEE Chemistry 2024 Question with Solution

The mass of sodium acetate (CH3COONa\text{CH}_3\text{COONa}) required to prepare 250 mL250 \, \text{mL} of 0.35 M0.35 \, \text{M} aqueous solution is (in grams). (Molar mass of CH3COONa\text{CH}_3\text{COONa} is 82.02 g/mol82.02 \, \text{g/mol})

Answer

Correct answer:7.18

Step-by-step solution

Standard Method

Given: Molarity of solution is 0.35 mol/L0.35 \, \text{mol/L}, volume is 250 mL250 \, \text{mL}, and molar mass of sodium acetate (CH3COONa)\left(\text{CH}_3\text{COONa}\right) is 82.02 g/mol82.02 \, \text{g/mol}.

Find: The mass of sodium acetate required.

Use the molarity relation:

Moles=M×V\text{Moles} = M \times V

Convert volume into litres:

250 mL=0.250 L250 \, \text{mL} = 0.250 \, \text{L}

Now calculate the number of moles:

Moles=0.35×0.25=0.0875 mol\text{Moles} = 0.35 \times 0.25 = 0.0875 \, \text{mol}

Use the mass relation:

Mass=moles×molar mass\text{Mass} = \text{moles} \times \text{molar mass}

Substituting the values:

Mass=0.0875×82.02=7.17675 g\text{Mass} = 0.0875 \times 82.02 = 7.17675 \, \text{g}

Therefore, the mass of sodium acetate required is 7.18 g7.18 \, \text{g}.

Detailed Computation

Given: M=0.35 MM = 0.35 \, \text{M}, V=250 mLV = 250 \, \text{mL}, and molar mass =82.02 g/mol= 82.02 \, \text{g/mol}.

Find: Required mass of CH3COONa\text{CH}_3\text{COONa}.

Molarity is defined as:

M=moles of solutevolume of solution in LM = \frac{\text{moles of solute}}{\text{volume of solution in L}}

So,

moles of solute=M×V\text{moles of solute} = M \times V

Convert the given volume to litres:

V=250 mL×1 L1000 mL=0.250 LV = 250 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.250 \, \text{L}

Calculate moles of sodium acetate:

moles=0.35 mol/L×0.250 L=0.0875 mol\text{moles} = 0.35 \, \text{mol/L} \times 0.250 \, \text{L} = 0.0875 \, \text{mol}

Now use the relation between mass, moles, and molar mass:

mass=moles×molar mass\text{mass} = \text{moles} \times \text{molar mass} mass=0.0875 mol×82.02 g/mol=7.17675 g\text{mass} = 0.0875 \, \text{mol} \times 82.02 \, \text{g/mol} = 7.17675 \, \text{g}

The solution also shows an approximate value of 7 g7 \, \text{g}, but the worked calculation gives 7.18 g7.18 \, \text{g}. Therefore, the numerical answer from the solution working is 7.187.18.

Common mistakes

  • Using 250250 directly as litres is incorrect because molarity requires volume in litres. Convert 250 mL250 \, \text{mL} to 0.250 L0.250 \, \text{L} before substitution.

  • Dividing by molar mass instead of multiplying after finding moles is wrong. Once moles are known, mass is obtained from mass=moles×molar mass\text{mass} = \text{moles} \times \text{molar mass}.

  • Rounding too early can change the final numerical answer. First calculate the mass as 7.17675 g7.17675 \, \text{g}, then round at the end.

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