Answer:
Molarity = 0.5 M
Explanation:
Given data:
Mass of NaCl = 2.7 g
Volume = 100 mL(100×10⁻³L)
Molarity of solution = ?
Solution:
Formula:
Molarity = number of moles / volume in litter
Number of moles:
Number of moles = mass/ molar mass
Number of moles = 2.7 g/ 58.5 g/mol
Number of moles = 0.05 mol
Molarity = number of moles / volume in litte
Molarity = 0.05 mol / 100×10⁻³L
Molarity = 0.5 M
Answer:
1023.75mmHg
Explanation:
V1 = 3.5L
P1 = 585mmHg
V2 = 2.0L
P2 = ?
To solve this question, we'll require the use of Boyle's law which states that the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature is kept constant.
Mathematically,
V = kP, k = PV
P1 × V1 = P2 × V2 = P3 × V3 = .......= Pn × Vn
P1 × V1 = P2 × V2
Solve for P2,
P2 = (P1 × V1) / V2
P2 = (585 × 3.5) / 2.0
P2 = 2047.5 / 2.0
P2 = 1023.75mmHg
The final pressure of the gas is 1023.75mmHg
1,g 2b 3c 4a 6e 7f !!!!!!!!!!!!
Answer:
Molarity = 0.809 M
mole fraction = 0.047
Explanation:
The complete question is
Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Solution -
Solution for molarity:
1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.
1)
Mass of 1.09 mole of acetone
= 1.09 mol x 58.0794 g/mol = 63.306 g
Density of acetone = 0.788 g/cm3
Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3
For ethanol
1000 g divided by 0.789 g/cm3 = 1267.427 cm3
Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3 = 1.347 L
a) Molarity:
1.09 mol / 1.347 L = 0.809 M
Mole Fraction
a) moles of ethanol:
1000 g / 46.0684 g/mol = 21.71 mol
b) moles of acetone:
1.09 / (1.09 + 21.71) = 0.047
I believe the mass of the product will be 400 grams. This is because according to the law of conservation of mass, mass is neither created nor destroyed . In a closed system, the mass of reactants is equal to mass of products. Therefore, since the total mass of the reactants is 400 grams then the mass of the products will still be 400 grams
