A) 70 amu
B) 60 amu
C) 50 amu
D) 40 amu
Correct Answer: C
Solution :
\[A+2B+3CA{{B}_{2}}{{C}_{3}}\]give: 6.0 g of \[A,6.0\times {{10}^{23}}\] atoms of B and 0.036 mole of C yields 4.8 gm of compound \[A{{B}_{2}}{{C}_{3}}.\] Atomic mass of A = 60 amu Atomic mass of C = 80 amu Mole of \[A=\frac{6}{60}=\frac{1}{10}=0.1\]mole Mole of \[B=\frac{6.0\times {{10}^{23}}}{6.023\times {{10}^{23}}}=1\]mole Mole of C = 0.036 So according to reaction \[A+2B+3CA{{B}_{2}}{{C}_{3}}\] C is limiting reagent which consumed \[=\frac{0.036}{3}\Rightarrow 0.22\]mole So 0.012 mole of C formed 0.012 mole of \[A{{B}_{2}}{{C}_{3}}.\] So Mole of \[A{{B}_{2}}{{C}_{3}}=\frac{\text{wt}}{\text{molecular}\,\text{wt}}\] \[0.012=\frac{4.8}{\text{Molecular}\,\text{wt}}\]of \[A{{B}_{2}}{{C}_{3}}\] So Molecular wt. of \[A{{B}_{2}}{{C}_{3}}=400\] So atomic mass of \[A+2\times \] Atomic mass of \[B+3\]atomic mass of \[C=400\] \[60+2B+3\times 80=400\] So atomic mass of \[B=50\] amu
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