Question

250 mL of 0.5 M $NaOH$ was added to 500 mL of 1M $HCl$.The number of unreacted $HCl$ molecules in the solution after compelete reaction is $x\times {10}^{21}$. The value of x to the nearest integer is
Take
($N_A=6.022\times {10}^{23}$)

Answer 1

$\text{Millimoles of}NaOH=250\times 0.5=125$

$\text{Millimoles of}HCl=500\times 1=500$
Now reaction is
$NaOH+HCl\to NaCl+H_{2}Ot=0125500--t=t0375125125$
So millimoles left of $HCl=375$
Moles of $HCl=375\times {10}^{-3}$
No.of HCl molecules
$=6.022\times {10}^{23}\times 375\times {10}^{-3}$
$=225.8\times {10}^{21}$
$\approx 226\times {10}^{21}$
value of x is 226