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NEET-XII-Chemistry

02: Solutions

with Solutions - page 4
Qstn# B-30 Prvs-QstnNext-Qstn
  • #30
    Calculate
    the amount of benzoic acid (C6H5COOH)
    required for preparing 250 mL of 0.15 M solution in methanol.
    Ans : 0.15 M solution of
    benzoic acid in methanol means,
    1000 mL of solution
    contains 0.15 mol of benzoic acid
    Therefore, 250 mL of
    solution contains = mol of benzoic acid
    = 0.0375 mol of
    benzoic acid
    Molar
    mass of benzoic acid (C6H5COOH)
    = 7 × 12 + 6 × 1 + 2 × 16
    =
    122 g mol-1
    Hence, required benzoic
    acid = 0.0375 mol × 122 g mol-1
    = 4.575 g