A) 445 g
B) 490 g
C) 495g
D) 890 g
Correct Answer: C
Solution :
Moles of \[{{C}_{57}}{{H}_{10}}{{O}_{6}}=\frac{445}{890}=0.5\] \[\frac{{{n}_{{{C}_{57}}{{H}_{110}}{{O}_{6}}}}}{2}=\frac{{{n}_{{{H}_{2}}O}}}{110}\] \[{{n}_{{{H}_{2}}O}}\,\,=55\,\,\times \,\,0.5\,\,\times \,\,18=495\,gm\]You need to login to perform this action.
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