Calculate the molality ( m) of solution containing 30.5 g of ethylene glycol C2H6O2 dissolved in 185 g of water. SHOW WORK PLEASE

GIVEN

mass of solution (m) =30.5g

mass of solvent (ms) =185g

R.tc Malality (M) =?

sln

mr=C2H6O2

mr=12x2+1x6+16x2

=62

Malality =mX1000 /mrxms

M =30.5x1000/62x185

Malality =2.659

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given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g