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What is the normality of lead (II) nitrate if the density of its 26% (w/w) aqueous solution is 3.105 g/mL? Take molar mass of lead (II) nitrate to be 331g/mol.(a) 2.437 N(b) 4.878 N(c) 0.243 N(d) 0.488 NThis question was posed to me in an online quiz.The doubt is from Expressing Concentration of Solutions in division Solutions of Chemistry – Class 12

Answer»

The correct option is (B) 4.878 N

For explanation: CONSIDER 100g of solution. It is made up of 26g lead (II) NITRATE and 74g water.

Volume of solution, V = 100g/(3.105g/ml) = 32.2061 ml = 0.0322 L

Equivalent weight of lead nitrate = 331/2 = 165.5 g/eq

Number of equivalents, N = 26g/(165.5g/eq) = 0.1571 eq

Normality = N/V = 4.878 N.



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