$ 1 \mathrm{M} \mathrm{NaOH} $ solution was slowly added into $ 1000 \mathrm{~mL} $ of $ 183.75 \mathrm{~g} $ impure $ \mathrm{H}_{2} \mathrm{SO}_{4} $ solution and the following plot was obtained. The percentage purity of $ \mathrm{H}_{2} \mathrm{SO}_{4} $ sample and slope of the curve respectively are :

Question

$ 1 \mathrm{M} \mathrm{NaOH} $ solution was slowly added into $ 1000 \mathrm{~mL} $ of $ 183.75 \mathrm{~g} $ impure $ \mathrm{H}_{2} \mathrm{SO}_{4} $ solution and the following plot was obtained. The percentage purity of $ \mathrm{H}_{2} \mathrm{SO}_{4} $ sample and slope of the curve respectively are :, $ 75 \%,-\frac{1}{3} $, $ 80 \%,-\frac{1}{2} $, $ 80 \%,-1 $, None of these

MARKS App · Accepted Answer

2 N a O H + H 2 S O 4 → N a 2 S O 4 + 2 H 2 O 2 Mole of NaOH react with 1 mole of H 2 SO 4. From given graph, Moles of NaOH consumed when reacted with H + = 3 × 1 N 1 V 1 = N 2 V 2 3L x 1M NaOH = 2 x 1L x MH 2 SO 4 moles of H 2 SO 4 present in sample = 1.5 wt. of H 2 SO 4 in sample = 1.5 × 98 = 147 g % purity = 1 4 7 183.75 × 1 0 0 = 8 0

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