A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
The answer is B: chronological order
V ( NaOH ) = mL ?
M ( NaOH ) = 0.100 M
V ( HCl ) = 9.00 mL / 1000 => 0.009 L
M ( HCl ) = 0.0500 M
number of moles HCl:
n = M x V
n = 0.009 x 0.0500 => 0.00045 moles HCl
mole ratio:
<span>HCl + NaOH = NaCl + H2O
</span>
1 mole HCl ---------------- 1 mole NaOH
0.00045 moles HCl ----- ??
0.00045 x 1 / 1 => 0.00045 moles of NaOH
M = n / V
0.100 = 0.00045 / V
V = 0.00045 / 0.100
V = 0.0045 L
1 L ------------ 1000 mL
0.0045 L ----- ??
0.0045 x 1000 / 1 => 4.5 mL of NaOH
Answer: Option (D) 30N
Detailed Solution:
According to Newton's second law:
F = ma --- (A)
Given:
mass = 5kg
acceleration = 6 m/s^2
F = ?
Plug all the value in equation (A)
F = (5)(6)
Ans: F = 30N
