Answer:
A. 0.433 moles of HNO₃
B. 6.07×10⁻⁵ moles of NaCl
C. 3.51×10⁻³ moles of sucrose
Explanation:
Part A.
Molarity . volume = moles
1.70 mol/L . 0.255L = 0.433 moles
Part B.
1.35 m of NaCl means, a molal concentration (m). Moles of solute in 1kg of solvent.
Let's convert 1 kg to mg for the rule of three
1 kg = 1×10⁶ mg
So in 1×10⁶ mg of solvent, we have 1.35 moles of solute (NaCl)
In 45 mg, we would have ( 45 . 1.35) / 1×10⁶ = 6.07×10⁻⁵ moles of NaCl
Part C. 1.50 % by mass means, 1.50 g of solute, in 100g of solution.
Let's make a rule of three:
In 100 g of solution we have 1.50 g of sucrose
In 80 g of solution, we would have (80 . 1.50) / 100 =1.2 g of sucrose.
Then, let's convert the mass in moles ( mass / molar mass)
1.2 g / 342 g/m = 3.51×10⁻³ moles
(A) 0.433 moles of HNO₃ in 255mL of 1.70M HNO3
(B) 6.07×10⁻⁵ moles of NaCl in 45.0mg of an aqueous solution that is 1.35m NaCl
(C) 3.51×10⁻³ moles of sucrose in an aqueous solution that is 1.50% sucrose (C12H22O11) by mass
(A) 255mL of 1.70M HNO3(aq)
Molarity × volume = moles
1.70 × 0.255L = 0.433 moles of HNO3
(B) 45.0mg of an aqueous solution that is 1.35m NaCl
1.35 m of NaCl means, a molal concentration (m). Moles of solute in 1kg of solvent.
In 45 mg, the number of moles is:
45 × 1.35 / 1×10⁶ = 6.07×10⁻⁵ moles of NaCl
(C) 80.0g of an aqueous solution that is 1.50% sucrose (C12H22O11) by mass
the solute is 1.50 % by mass, that is :
1.50 g of solute, in 100g of solution.
In 80 g of solution
80 × 1.50 / 100 =1.2 g of sucrose.
Now, convert the mass in moles
number of moles = given mass / molar mass
1.2 g / 342 g/m = 3.51×10⁻³ moles of sucrose
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