Answer: 41.5 mL
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,
n = moles of solute
= volume of solution in L
Given : 59.4 g of
in 100 g of solution
moles of 
Volume of solution =
Now put all the given values in the formula of molality, we get

To calculate the volume of acid, we use the equation given by neutralisation reaction:

where,
are the molarity and volume of stock acid which is 
are the molarity and volume of dilute acid which is 
We are given:

Putting values in above equation, we get:

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid
Answer:
Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)
Explanation:
Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)
This is how it starts out.
Left:
Right
So the place to start with this equation is to bring the Cls up to 2
Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)
But the Nas are now out of kilter.
Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)
Now the right has a problem. There's only 1 Na
Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)
Check it out. It looks like we are done.