Answer:
0.324 g is required to make 5.00 M solution of NaCl in 0.800 L.
Given data:
Molarity = 5.00 M
Formula Mass = 58.5 g/mol
Required volume = 0.800 L
To Find;
Mass in gram = ?
Solution:
Formula for calculating mass in gram is given as,
Mass in gram = Molarity × Formula mass × Volume required / 1000 putting values
Mass in gram = 5.00 M × 58.5 g/mol × 0.800 L / 1000
Mass in gram = 0.234 g
First, we write the balanced equation for this reaction:
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:
Moles = volume (in L) * molarity
We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:
M₁V₁ = 2M₂V₂
V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L
The volume required is 173.6 mL
Remembering that
d = m ÷ v
d = ?
m = 89 g
v = 10 cm³
Therefore:
d = 89 ÷ 10
d = 8,9 g÷cm³
Astatine. Because it has the smaller shell of electrons. I believe