Question

the density of NH4OH solution is 0.6 g/ml. It contains 34% by weight of NH4OH. Calculate the normality of the solution.

Answer 1

Hi friend,

# 17.64 N

Solution :-

In 1ml of solution we have 0.6g NH4OH.

But the defenation of normality says moles in 1 L solution.

So, the mass of solution ( NH4OH) in 1,000 ml = 0.6×1000ml = 600g

Thus , moles = mass / equivalent mass

Equivalent mass = Molar mass / n-factor

n-factor for NH4OH is 1

So, 600/34 = 17.64 N (proved )...i hope it helps you

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Answer 2

Answer:  The normality of solution is  4.35 N.

Solution : Given,  34% by weight of $NH_4OH$ means 34 g of $NH_4OH$ are present in 100 g of water.

given mass of $NH_4OH$ = 34 g

Mass of water (solvent) = 100 g

Mass of solution = Mass of solute + Mass of solvent = 34 g + 100 g  = 134 g

Now we have to calculate the volume of solution.

$Volume=\frac{Mass}{Density}=\frac{134g}{0.6g/ml}=223.3ml=0.223L$  

Normality : It is defined as the number of gram equivalents of solute present in one liter of solution.

Formula used :

$Normality=\frac{n}{V_s}$

where,

n = gram equivalents

$V_s$ = volume of solution in liter

${\text{Equivalent mass}}=\frac{\text{Molecular mass}}{\text{Acidity}}$

$\text{Equivalent mass of }NH_4OH=\frac{35}{1}=35$

${\text{gram equivalents}}=\frac{\text{Given mass}}{\text{Equivalent mass}}$

${\text{gram equivalents}}=\frac{34}{35}=0.97$

$Normality=\frac{0.97}{ 0.223L}=4.35N$

Therefore, the Normality of solution will be 4.35N.