A) \[13.6g,\text{ }0.28\text{ }mol\text{ }{{L}^{-1}}\]
B) \[13.6\,g,\text{ }0.14\,mol\,{{L}^{-\,1}}\]
C) \[1.9\,g,\text{ }0.28\text{ }mol\text{ }{{L}^{-1}}\]
D) \[1.9\,g,\text{ }0.14\text{ }mol\,{{L}^{-1}}\]
Correct Answer: C
Solution :
\[Ca{{\left( OH \right)}_{2}}\text{+}N{{a}_{2}}S{{O}_{4}}\to CaS{{O}_{4}}+2NaOH\] 100 m mole 14 m mole 86 m mole - 14 m mole 28 m mole \[\therefore \text{ }W\text{ }CaS{{O}_{4}}=14\times {{10}^{-3}}\times 136\] \[=\text{ }1.9\text{ }gm\] \[[O{{H}^{-}}]=\frac{28}{100}\,\,=\,\,\,0.28\,M\]You need to login to perform this action.
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