Question #3dbe1
1 Answer
Explanation:
A. Assuming that the concentration of the solution presented is in %w/w; thus the formula shown below is applicable to solve this problem.
#%solution(w/w)=("mass solute"(m))/("mass solution"(m_s))#
where:
#% solution =17%#
#"mass solute"=35gNaCl#
#rho " solution"=(1.13g)/(cm^3)# Now, plug in given data to its respective variable to get the intermediate value needed to solve the prescribed problem. Rearrange formula as needed to isolate the required variable; that is,
#%solution(w/w)=m/m_s#
#m_s=(m)/(%solution)#
#m_s=(35g)/(0.17)#
#m_s=205.8824g# Knowing the
#m_s# , the total volume of the solution that also contained the given mass of the solute can be calculated through the density formula; thus,
#rho_s=m_s/V_s#
where:
#rho_s="density of the solution"#
#m_s="mass of the solution"#
#V_s="volume of the solution"#
#V_s=m_s/rho_s#
#V_s=(205.8824cancel(g))/((1.13cancel(g))/(ml))#
#V_s~~183ml#
B. Assuming that the concentration of the solution presented is again in %w/w; thus the formula shown below is applicable to solve this problem.
#%solution(w/w)=("mass solute"(m))/("mass solution"(m_s))#
where:
#% solution =58%#
#"mass solute"=150g H_2SO_4# Now, plug in given data to its respective variable to get the intermediate value needed to solve the prescribed problem. Rearrange formula as needed to isolate the required variable; that is,
#%solution(w/w)=m/m_s#
#m_s=(m)/(%solution)#
#m_s=(150g)/(0.58)#
#m_s=260g#
