JEE Main 2015 · 11 Apr · Shift 1 · Chemistry Q31

← JEE Main 2015 · 11 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 31 of 90

$\ce{A + 2B + 3C <=> AB2C3}$

Reaction of 6.0 g of A, $6.0\times10^{23}$ atoms of B, and 0.036 mol of C yields 4.8 g of compound $\ce{AB2C3}$ . If the atomic mass of A and C are 60 and 80 amu, respectively, the atomic mass of B is (Avogadro no. = $6\times10^{23}$ ) :

70 amu

60 amu

50 amu

40 amu

Answer and worked solution

Official answer: C

Moles: A 0.1, B 1.0, C 0.036. C is limiting and gives $\frac{0.036}3=0.012$ mol of product.
$M=\frac{4.8}{0.012}=400=60+2M_B+3(80)\implies M_B=50\text{ amu}.$

Topics:Stoichiometry

← Previous question Next question →

JEE Main 2015 · 11 Apr · Shift 1, Chemistry question 31