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Q)

The molarity of con.HCl is $1.15\;M$. What volume of con. HCl is required to make 1.00 of 0.1M HCl

$\begin{array}{1 1} (a) 56.9\;ml\\ (b) 89\;ml\\ (c) 86.9\;ml\\ (d) 80\;ml \end{array} $

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A)
Solution :
$V_1M_1=V_2M_2$
$1000\;ml \times 0.1\;M =V_2 \times 1.15\;M$
$V_2=\large\frac{1000 \times 0.1}{1.15}\;ml$
$\qquad= \large\frac{100}{1.15}$
$\qquad= \large\frac{100 \times 10^{2}}{115}$$=86.9\;ml$
 
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A)
Solution : 86.9 ml
 

 

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