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
The third question requires you to solve for the weight of sodium (Na) and weight of Chloride (Cl) from the calculated moles of each element Na, and Cl.
So, you need to multiply the calculated moles of Na with its molar mass (23 g/ mol) to get the answer for Na. And multiply the calculated moles of Cl with its molar mass (35.45 g/mol) to get the answer for Cl.
The answer is 4.9 moles.
Solution:
Using the equation for boiling point elevation Δt,
Δt = i Kb m
we can rearrange the expression to solve for the molality m of the solution:
m = Δt / i Kb
Since we know that pure water boils at 100 °C, and the Ebullioscopic constant Kb for water is 0.512 °C·kg/mol,
m = (105°C - 100°C) / (2 * 0.512 °C·kg/mol)
= 4.883 mol/kg
From the molality m of the solution of salt added in a kilogram of water, we can now find the number of moles of salt:
m = number of moles / 1.0kg
number of moles = m*1.0kg
= (4.883 mol/kg) * (1.0kg)
= 4.9 moles
Answer:
Calendar and periodic table both have repetitive patterns.
Explanation:
In calendar the days are arranged and divided into weeks whereas in the periodic table the elements are arranged in increasing atomic number and divided into groups
