Question

If the density of methanol is 0.793 kg L-1, what is its volume needed for making
2.5 L of its 0.25 M solution?
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Answer 1

⭐️ Required Solution ⭐️

Moles of methanol present in 2.5 L of 0.25 M solution

$molarity \: = \frac{molesofch ^{3}oh }{volumeinlitre}$

$0.25 = \frac{molesofch^{3}oh }{2.5}$

$moles \: of \: ch ^{3} oh = 2.5 \times 0.25 = 0.625$

$mass \: of \: ch^{3} oh = 0.625 \times 32 = 20g$

$(molecular \: mass \: of \: ch ^{3}oh = 12 + 1 \times 3 + 16 + 1 = 32$

$0.793 \times 10 ^{3}g \: of \: ch ^{3} oh \: is \: present \: in \: 1000 \: ml$

$20g \: of \: ch ^{3} oh \: is \: present \: in \: = \frac{1000}{0.793 \times 10 ^{3} } \times 20 = 25.2 \: ml$