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Q.
The volumes of 4 N HCl and 10 N HCl requiredto make 1 litre of 6 N HCI are
Solutions
Solution:
Let $N _{1}$ and $V _{1}$ be normality and volume of $4 N HCl$ and $N _{2}$ be normality and volume of $10 N HCl$ respectively.
As volume to make is $1 L$ and Normality to make is $6 N$,
$
\text { so } V _{1}+ V _{2}= V
$
As we know $N _{1} V _{1}+ N _{2} V _{2}= NV$
Calculation to determine $V _{1}$ and $V _{2}$ :
Substituting the values in $N _{1} V _{1}+ N _{2} V _{2}= NV$ equation
$
4 \times V_{1}+10 \times\left(1-V_{1}\right)=6 \times 1 \quad\left(\text { As } V_{2}=1-V_{1}\right)
$
so $V _{1}=2 / 3$ and $V _{2}=1 / 3$
$V _{1}=2 / 3=0 . .67$ litre and $V _{2}=0.33$ litre
Thus, answer is $0.67$ litre of $4 NHCl$ and $0.33$ litre of $10 NHCl$.