In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
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Answer:
0.302 moles
Explanation:
Data given
Mass of Pb(NO₃)₂ = 100 g
Moles of Pb(NO₃)₂ = ?
Solution:
To find mole we have to know about molar mass of Pb(NO₃)₂
So,
Molar mass of Pb(NO₃)₂ = 207 + 2[14 + 3(16)]
= 207 + 2[14 + 48]
= 207 + 124
Molar mass of Pb(NO₃)₂ = 331 g/mol
Formula used :
no. of moles = mass in grams / molar mass
Put values in above formula
no. of moles = 100 g / 331 g/mol
no. of moles = 0.302 moles
no. of moles of Pb(NO₃)₂ = 0.302 moles
Answer:
0.2 mL stock solution, 0.8 solvent, 0.1 mL first solution and 0.9 solvent
Explanation:
The final volume for fist solution is 1 mL and concentration must will be 1/5, then 1 mL/5=0.2 mL. For complete the 1 mL add the missing solvent volume 1 mL-0.2 mL=0.8 mL. For second solution, assuming final volume is 1 mL, and concentration 1/10, then we have 1 mL /10=0.1 mL solution 1/5. Completing volume, 1 mL-0.1 mL= 0.9 mL solvent.
