How many mL of a 1.63 M solution would contain 12 g of Na(OH)?
Answers
Answer:
Answer : 89 mL of solution would contain the given amount of Al(NO₃)₃.
Explanation :
Step 1 : Find moles of Al(NO₃)₃.
The molar mass of Al(NO₃)₃ is 213 g/mol
The formula to calculate mole is given below.
Mole = \frac{Mass (grams)}{MolarMass}Mole=
MolarMass
Mass(grams)
We have 12 g of Al(NO₃)₃. Let us plug in this value to find mol.
mole = \frac{12g}{213 g/mol}mole=
213g/mol
12g
Mole = 0.056 mol.
We have 0.056 mols of Al(NO₃)₃
Step 2 : Use molarity formula to find the volume.
The molarity of a solution is defined as moles of solute per liter of solution.
This can be represented in terms of formula as follows.
Molarity (M)= \frac{mol}{L}Molarity(M)=
L
mol
We have 0.63 M solution.
0.63 M = \frac{0.056mol}{L}0.63M=
L
0.056mol
On rearranging we get,
L = \frac{0.056}{0.63} = 0.089L=
0.63
0.056
=0.089
We have 0.089 L of solution. Let us convert this to mL.
0.089 L \times \frac{1000mL}{1L} = 89 mL0.089L×
1L
1000mL
=89mL
89 mL of solution would contain the given amount of Al(NO₃)₃
Answer:
184ml
Explanation:
M=n×1000/V
V=n×1000/M
=12000/40×1.63
=184.04