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 331 g/mol.

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0.243 N

0.488 N

2.437 N

4.878 N

answer is B.

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Detailed Solution

Consider 100 g of solution. It is made up of 26 g lead (II) nitrate and 74 g water.
Volume of solution,

V=100 g/(3.105 g/ml)=32.2061ml=0.0322 L

V=100 g/(3.105 g/ml)=32.2061ml=0.0322 L
Equivalent weight of lead nitrate

=331/2=165.5 g/eq=331/2=165.5 g/eq
Number of equivalents,

N=26 g/(165.5 g/eq)=0.1571eqN=26 g/(165.5 g/eq)=0.1571eq
Normality =N/V=4.878 N=N/V=4.878 N

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