A) 80%
B) 99%
C) 75%
D) 100% \[1.5\mu \] \[\mu \] Normal molecular weight of \[W\] \[\frac{4W}{3}\]=122 \[\frac{5W}{2}\] \[\frac{\pi }{2}\] \[\sigma =\text{5}.\text{67}\times \text{1}{{0}^{-\text{8}}}\text{W}-{{\text{m}}^{\text{2}}}{{\text{K}}^{\text{-4}}}\] \[2{{C}_{6}}{{H}_{5}}COOH\xrightarrow{{}}{{({{C}_{6}}{{H}_{5}}COOH)}_{2}}\] \[t\] where, \[{{(Kg)}^{1/2}}\] \[{{(Kg)}^{-1/2}}\] \[{{(Kg)}^{2}}\] % association \[{{(Kg)}^{-2}}\] \[\frac{pV}{nT}\]
Correct Answer: B
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