Answer:
1.811 g
Explanation:
The computation of the mass need to use to make the solution is shown below:
We know that molarity is

So,


= 0.031 moles
Now

where,
The Molecular weight of NaCl is 58.44 g/mole
And, the moles are 0.031 moles
So, the mass of NaCL is

= 1.811 g
We simply applied the above formulas
Answer is: <span>the mass of the glucose is 81,07 grams.
</span>c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.
Answer:

Explanation:
![\Delta H_{rxn}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28product%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28reactant_%7Bj%7D%29%5D)
Where
and
are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).
is standard heat of formation.
So, ![\Delta H_{rxn}=[2mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[3mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{2}H_{5}OH)_{l}]-[3mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B2mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B2%7DH_%7B5%7DOH%29_%7Bl%7D%5D-%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
or, ![\Delta H_{rxn}=[2mol\times -393.509kJ/mol]+[3mol\times -241.818kJ/mol]-[1mol\times -277.69kJ/mol]-[3mol\times 0kJ/mol]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B2mol%5Ctimes%20-393.509kJ%2Fmol%5D%2B%5B3mol%5Ctimes%20-241.818kJ%2Fmol%5D-%5B1mol%5Ctimes%20-277.69kJ%2Fmol%5D-%5B3mol%5Ctimes%200kJ%2Fmol%5D)
or, 
SI Unit
I'm not 100% sure but I think that's right. Hope I helped you in any way.