Question

A flask of volume 2.0dm³ was found to contain 5.28g. The pressure of the flask was 200 KPa at 20 degrees c temperature.calculate the relative molecular mass​

Answer 1

Explanation:

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Answer 2

Step by step explanation:

Given : A flask of volume $2.0dm^3$  was found to contain $5.28g$ . The pressure of the flask was $200 KPa$  at $20\°C$ temperature.

To find : The relative molecular mass

Formula used : We use Ideal gas equation,

⇒  $PV=nRT$

P is the pressure, V is the volume, n is the number of moles, R is universal gas constant and T is the temperature.

Since n is representing the no. of moles so we can write,

⇒  $n=\frac{W}{MM}$

W is given mass and MM is the molar mass of compound.

now the formula is ;

⇒  $PV=\frac{W}{MM} RT$

Conversion units

1). Pressure units :   $1KPa=1000Pa$

                                 $1Pa=\frac{1}{101325} atm$

2). Volume units :   $1dm^3=1L$

3). Temperature units :  $K=273.15+(\°C)$

• Calculation for molar mass,

We have values,

$P=200KPa$

$V=2.0 dm^3$

$W=5.28g$

$R=0.0821atmLK^{-1}mol^{-1}$

$T =20\°C$

Changing units of given data into standard units,

⇒ $P=200*10^3Pa$

        $=\frac{200*10^3}{101325}atm$

        $=1.974atm$

⇒  $V=2.0L$

⇒  $W=5.28g$

⇒  $R=0.0821atmLK^{-1}mol^{-1}$

⇒  $T=20+273.15K$

        $=293.15K$

Now using formula of ideal gas equation,

⇒  $PV=\frac{W}{MM} RT$

$(1.974atm)*(2.0L)=\frac{5.28g}{MM} *(0.082atmLK^{-1}mol^{-1})*( 293.15K)$

taking MM to other side,$MM=\frac{5.28g}{(1.974atm)*(2.0L)} *(0.082atmLK^{-1}mol^{-1})*( 293.15K)$

        $=\frac{5.28g}{3.948} *24.0383mol^{-1}$

       $=1.337g *24.0383mol^{-1}$

       $=32.14gmol^{-1}$

Hence the relative molecular mass is $32.14gmol^{-1}$.