Question

Which of the following statements is(are) correct?

[Atomic weight of $Bi=209$ g, the molecular weight of $Bi(N{O}_3{)}_3.5H_{2}O=485$ g/mol]

• A. $Bi+4HN{O}_3+3H_{2}O\to Bi(N{O}_3{)}_3.5H_{2}O+NO$
  $2.09$ g of $Bi$ in $HN{O}_3$ produces $48.5$ g of bismuth nitrate.
• B. $4.0$ g of $63\%$ $HN{O}_3$ by mass is required to react with $2.09$ g of $Bi$.
• C. The volume of $NO$ gas produced at STP ($1$ bar, $273$ K) is $0.227$ L.
• D. The volume of $NO$ gas produced at STP ($1$ bar, $298$ K) is $0.247$ L.

Answer 1

Answer: (B), (C), (D)

The correct options are
B $4.0$ g of $63\%$ $HN{O}_3$ by mass is required to react with $2.09$ g of $Bi$.
C The volume of $NO$ gas produced at STP ($1$ bar, $298$ K) is $0.247$ L.
D The volume of $NO$ gas produced at STP ($1$ bar, $273$ K) is $0.227$ L.

$\text{A}.$ Moles of $Bi=\frac{2.09}{209}=0.01mol$

Weight of bismuth nitrate $=(0.01$ mol Bi)

$\left(\frac{1molBi(N{O}_3{)}_3\cdot 5H_{2}O}{molBi}\right)\left(\frac{485gBi(N{O}_3{)}_3\cdot 5H_{2}O}{molBi(N{O}_3{)}_3\cdot 5H_{2}O}\right)$

$=0.01\times 484=4.85g$

Hence, (a) is wrong.

$\text{B}.$ Weight of $HN{O}_3=\left(0.01molBi\right)$

$\left(\frac{4molHN{O}_3}{molBi}\right)\left(\frac{63gHN{O}_3}{molHN{O}_3}\right)\left(\frac{100gsolution}{63gHN{O}_3}\right)$

$=\frac{0.01\times 4\times 63\times 100}{63}=4gHN{O}_3$

Hence, $\text{B}$ is correct.

$\text{C}.$ Molar volume of gas at STP (1 bar, 273 K)$=22.7L$

$\therefore$ Volume of NO gas $=0.01\times 22.7=0.227L$

Hence, $\text{C}$ is correct.

$\text{D}.$ Molar volume of gas at SATP (1 bar, 298 K)$=24.7L$

$\therefore$ Volume of NO gas $=0.01\times 24.7=0.247L$

Hence, $\text{D}$ is correct.

Hence options $\text{B,C}$ & $\text{D}$ are correct.