That way the solution is througly dissolve
Answer:
0.024M
Explanation:
Data obtained from the question include:
C1 = 6M
V1 = 2mL
C2 =?
V2 = 500mL
The molarity of the diluted solution can be obtained as follows:
C1V1 = C2V2
6 x 2 = C2 x 500
Divide both side by 500
C2 = (6 x 2) /500
C2 = 0.024M
The molarity of the diluted solution is 0.024M
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The answer is 62.00 g/mol.
Solution:
Knowing that the freezing point of water is 0°C, temperature change Δt is
Δt = 0C - (-1.23°C) = 1.23°C
Since the van 't Hoff factor i is essentially 1 for non-electrolytes dissolved in water, we calculate for the number of moles x of the compound dissolved from the equation
Δt = i Kf m
1.23°C = (1) (1.86°C kg mol-1) (x / 0.105 kg)
x = 0.069435 mol
Therefore, the molar mass of the solute is
molar mass = 4.305g / 0.069435mol = 62.00 g/mol
Answer:
The answer to your question is: 0.3 moles of AgNO₃
Explanation:
1.0 L sample
0.1 mol of NaCl
0.1 mol of CaCl₂
AgNO₃ = ? moles
Reactions
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
Then 1 NaCl mol --------------- 1 AgNO₃
0.1 mol -------------- x
x = 0.1 moles of AgNO₃ needed
CaCl₂ + 2 AgNO₃ ⇒ 2 AgCl + Ca(NO₃)₂
Then 1 mol of CaCl₂ ------------- 2 moles of AgNO₃
0.1 mol ------------- x
x = 0.2 moles of AgNO₃
Total moles of AgNO₃ = 0.1 + 0.2 = 0.3
