Question

QUESTION

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How much volume of 98% by mass H2SO4 [density 1.8 gram per ml] required to prepare 500 ml of 23 % by mass
H2 SO4 [density 1.2 gram per ml]


Also, calculate the mass of H2O added?


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HARSH PRATAP SINGH ​

Answer 1

Molar mass of H2SO4 = 98g/mol = Mm

D =density = 1.84g/cc

W/w = 98%

V = 1dm3 = 1L = 1000ml = 1000cc

Mass of solution = V x D

M = 1000 x 1.84 = 1840g per 1000cc

Using w/w, 98% of 1840g = H2SO4

= (98/100) x 1840 = 1803.2g = m

Nos of moles = m/Mm = 1803.2/98 = 18.4moles per 1000cc

Thus, molarity = 18.4M (18.4mol/dm3 )

Answer 2

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$98\% \: by \: \: weight \: \: of \: \: solution \\ means \: 98g \: of \: H_2SO_4 \: is \: present \: in \: 100g \: of\: solution \\ \\ volume \: of \: solution = \frac{mass \: of \: solution}{density \: of \: solution} \\ \\ = \frac{100g}{(1.8 \times 1000g) L</p><p>} \\ \\ volume \: of \: solution = \frac{1}{18} L \\ \\ \% \: H_2SO_4 \: \: w/v \: \: of \: solution = molarity \\ \\ = \frac{no \: of \: moles \: of \:H_2SO_4 }{volume \: of \: solution \: (L)} \\ \\ molarity = \\ \frac{given \: mass \: of \:H_2SO_4 / \: molar \: mass \: of \:H_2SO_4 }{volume \: of \: solution \: } \\ \\ molarity = \frac{98/98}{1/18} = 18 M$