Answer:
a. 5.77L
b. 700mmHg
c. 395K
Explanation:
Using PV = nRT we can solve these problems where:
P is pressure of the gas in atm (1atm = 760mmHg)
V is volume in liters
n are moles of the gas
R is gas constant: 0.082atmL/molK
T is asbolute temperature in K
a. PV = nRT
V = nRT/P
P = 773mmHg*(1atm/760mmHg) = 1.017atm
T = 25°C+273 = 298K
V = 0.240mol*0.082atmL/molK*298K / 1.017atm
V = 5.77L
b. PV = nRT
P = nRT/V
P = 0.0947mol*0.082atmL/molK*309K/0.635L
P = 0.9216atm * (760mmHg/1atm) = 700mmHg
c. PV = nRT
PV/nR = T
P = 727mmHg * (1atm / 760mmHg) = 0.9566atm
0.9566atm*13.3L/0.393mol*0.082atmL/molK = T
T = 395K
Answer:
The solution would need 13.9 g of KCl
Explanation:
0.75 m, means molal concentration
0.75 moles in 1 kg of solvent.
Let's think as an aqueous solution.
250 mL = 250 g, cause water density (1g/mL)
1000 g have 0.75 moles of solute
250 g will have (0.75 . 250)/1000 = 0.1875 moles of KCl
Let's convert that moles in mass (mol . molar mass)
0.1875 m . 74.55 g/m = 13.9 g
Explanation:
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1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution
=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution
2) Molarity = M = number of moles of solute / liters of solution
=> calculate the number of moles of 8.9 grams of NaCl
3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol
4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol
5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M
Answer: 0.152 M
