A gaseous mixture containing 1.5 mol Ar and 3.5 mol CO_2 has a total pressure at

Question

A gaseous mixture containing 1.5 mol Ar and 3.5 mol CO_2 has a total pressure atm. What is the partial pressure of CO_2? 1.8 atm 2.1 atm 3.5 atm 2.4 atm 4.9 atm Calculate the density of oxygen (O_2) at STP. 0.70 g/mL 1.43 g/mL 70.9 g/mL 0.76g/mL None of the above Which of the following is a strong acid in aqueous solution? H_2CO_3 H_2SO_4 H_2SO_3 H_3PO_4 None of the above With what volume of 5.0 M HF will 7.4 g of calcium hydroxide (M. Mass = g/mol) react completely, according to the reaction: 2HF + Ca(OH)_2 rightarrow CaF_2 + 50. mL 30. mL 20. mL 40. mL 1.0 times 10^2 mL

Explanation / Answer

Ans. 10. Mole fraction of CO2 = Number of moles of CO2 / total number of moles

                                    = 3.5 moles / (3.5 + 1.5) moles = 3.5 moles/ 5.0 moles

                                    = 0.7

Partial pressure of CO2 = mole fraction of CO2 x total pressure

                                    = 0.7 x (total pressure)

            Note: value of total pressure is not visible in text. Just multiply total pressure wo=ith 0.7 to get partial pressure of CO2

Ans. 11. Volume of 1 mole O2 at STP = 22.4141 L

Mass of 1 mole O2 gas = moles x molecular mass

                                    = 1 mole x 32.0 g mol-1

                                    = 32.0 g

Density = mass / volume

            = 32.0 g / 24.4141 L = 1.31 g/ L = 1.31 x 10-3 g/ mL = 0.00131 g / mL

Note: 1 L = 1000 mL

Correct option: none    

Ans. 12. B. H2SO4        

Ans, 13.

Number of moles of Ca(OH)2 = mass / molecular mass

                                                = 7.4 g / 74.0 g mol-1 = 0.10 moles

2 moles of HF completely react with 1 mol Ca(OH)2. For complete reaction, the total number of moles of Ca(OH)2 must be equal to (2 x number of moles) of HF.

                        Using, M1V1 = M2V2     --- equation 1

Where, M1= molarity of solution 1, V1= volume of solution 1 in Liters (HF solution)

Molarity = moles / L ; so Molarity x volume in L = number of moles

Or, M1V1 = 2 x number of moles of Ca(OH)2

Or, 5.0 M x V1 = 2 x 0.10 = 0.20

Or, V1 = 0.20 / 5 = 0.040 L

Hence, volume of HF = 0.040 L = 40.0 mL

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