N=m(g)/m.wt
n=85/12(1)+16(2) =1.93 moles
Answer:
1.0 M
Explanation:
Reaction equation;
KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)
Concentration of acid CA = ?
Concentration of base CB = 1.0 M
Volume of base VB = 25.60 - 0.50 = 25.1 ml
Volume of acid VB = 25.0 ml
Number of moles of acid NA = 1
Number of moles of base NB =2
CAVA/CBVB =NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 1 * 25.1 * 1/25.0 *1
CA = 1.0 M
The chemical properties of alcohol are compounds that possess a hydroxyl group (OH) connected an sp3 hybridized carbon hope i it help
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
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