A) 0.80 M
B) 1.0 M
C) 0.73 M
D) 0.50 M
Correct Answer: C
Solution :
| Moles of 2.5 L of 1 M \[NaOH=2.5\times 1=2.5\] |
| Moles of 3.0 L of 0.5 M \[NaOH=3.0\times 0.5=1.5\] |
| Total moles of NaOH in solution \[=2.5+1.5=4.0\] |
| (Total volume of solution = 2.5 + 3.0 = 5.5 L) |
| Thus \[{{M}_{1}}\times {{V}_{1}}={{M}_{2}}\times {{V}_{2}}\] |
| \[4.0={{M}_{2}}\times 5.5\] |
| \[\therefore \] molarity of resultant \[t\] solution |
| \[={{M}_{2}}=\frac{4.0}{5.5}M\] |
| \[\approx \,\,0.73\,\,M\] |
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