Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
Unit of M is also mole/L, where mole is the moles of solute and L is the volume of the solution. The latter is given: 158 mL or 0.158 L. So we need to find out the moles of NH4Br.
Moles of NH4Br = Mass of NH4Br/molar mass of NH4Br = 17.0g/(14+1*4+79.9)g/mol = 0.1736 mole.
So, the molarity of the solution = 0.1736mole/0.158L = 1.10 mole/L = 1.10 M