112.0 mL of NO2 at STP was liquefied, the density of the liquid being 1.15 g mL-1. Calculate the volume and the number of molecules in the liquid NO2.

Question

112.0 mL of NO2 at STP was liquefied, the density of the liquid being 1.15 g mL-1. Calculate the volume and the number of molecules in the liquid NO2. (A) 0.10 mL and 3.01 × 1022 (B) 0.20 mL and 3.01 × 1021 (C) 0.20 mL and 6.02 × 1023 (D) 0.10 mL and 6.02 × 1021

Tardigrade Admin · Accepted Answer

At STP 22400 mL of NO2 â¡ 46 g of NO2 ∴ 112.0 mL of NO2=(112.0 mL×46.0 g/22400 mL)=0.23 g. VNO2=(mass/density)=(0.23/1.15 g mL-1)=0.20 mL. Number of molecules =(0.23/46)×6.02×1023=3.01×1021

Q. $112.0 m L$ of $N O_{2}$ at $STP$ was liquefied, the density of the liquid being $1.15 g m L^{- 1}$ . Calculate the volume and the number of molecules in the liquid $N O_{2} .$

$0.10 m L$ and $3.01 \times 1 0^{22}$

$0.20 m L$ and $3.01 \times 1 0^{21}$

$0.20 m L$ and $6.02 \times 1 0^{23}$

$0.10 m L$ and $6.02 \times 1 0^{21}$

Q. 112.0mL of NO2​ at STP was liquefied, the density of the liquid being 1.15gmL−1. Calculate the volume and the number of molecules in the liquid NO2​.

0.10mL and 3.01×1022

0.20mL and 3.01×1021

0.20mL and 6.02×1023

0.10mL and 6.02×1021

At STP22400mL of NO2​≡46g of NO2​ ∴112.0mLofNO2​=22400mL112.0mL×46.0g​=0.23g. VNO2​​=densitymass​=1.15gmL−10.23​=0.20mL. Number of molecules =460.23​×6.02×1023=3.01×1021

Q. $112.0 m L$ of $N O_{2}$ at $STP$ was liquefied, the density of the liquid being $1.15 g m L^{- 1}$ . Calculate the volume and the number of molecules in the liquid $N O_{2} .$

$0.10 m L$ and $3.01 \times 1 0^{22}$

$0.20 m L$ and $3.01 \times 1 0^{21}$

$0.20 m L$ and $6.02 \times 1 0^{23}$

$0.10 m L$ and $6.02 \times 1 0^{21}$