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Some Basic Concepts Of Chemistry

Question
CBSEENCH11005111
Wired Faculty App

If the density of methanol is 0·793 kg L–1, what is its volume needed for making 2·5L of its 0·25 M solution?

Solution

Molar mass of methanol (CH3OH)
                = 12 + 4 x 1 + 16 = 32g mol-1
                = 0.032 kg mol-1
therefore       Molarity of the given solution
       equals space fraction numerator 0.793 space kg space straight L to the power of negative 1 end exponent over denominator 0.032 space kg space mol to the power of negative 1 end exponent end fraction space equals space 24.78 space mol space straight L to the power of negative 1 end exponent
Applying the molarity equation,
(Given solution) straight M subscript 1 straight V subscript 1 space equals space straight M subscript 2 straight V subscript 2 left parenthesis solution space to space be space prepared right parenthesis
24.78 space cross times space straight V subscript 1 space equals space 0.25 space cross times space 2.5 space straight L
or     straight V subscript 1 space equals space fraction numerator 0.25 space cross times space 2.5 over denominator 24.78 end fraction straight L space equals space 0.02522 space straight L space equals space 25.22 space mL

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