Calculate the grams of solute needed to prepare each of the following solutions.See screenshot 1.8 L of a 4.5 M NAOH solutionExpress your answer using two significant figures.Π ΑΣφ?g NaOHm =SubmitRequest AnswerPart B6.5 L of a 0.50 M CaCl2 solutionExpress your answer using two significant figures.NV ΑΣφ?g CaCl2m =

Answered: 1.8 L of a 4.5 M NAOH solution Express…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Calculate the grams of solute needed to prepare each of the following solutions.

See screenshot

1.8 L of a 4.5 M NAOH solution
Express your answer using two significant figures.
Π ΑΣφ
?
g NaOH
m =
Submit
Request Answer
Part B
6.5 L of a 0.50 M CaCl2 solution
Express your answer using two significant figures.
NV ΑΣφ
?
g CaCl2
m =

Expert Solution

Step 1

$w e m u s t k n o w t w o b a s i c f o r m u l a e t o s o l v e t h e s e q u e s t i o n s : n u m b e r o f m o l e s o f s o l u t e = \frac{g i v e n m a s s o f s o l u t e}{m o l a r m a s s o l f s o l u t e} n_{s o l u t e} = \frac{m_{s o l u t e}}{M_{s o l u t e}} . . . . . . . . . . 1 s t . C o n c e n t r a t i o m / m o l a r i t y (M) = \frac{n o . o f m o l e s o f s o l u t e}{v o l u m e o f s o l u t i o n i n l i t e r s} M = \frac{n_{s o l u t e}}{V_{s o l u t i o n}} . . . . . . . . . . . . . . . . . 2 n d (a) 1.8 L o f 4.5 M N a O H s o u l t i o n V = 1.8 L M = 4.5 M M_{N a O H} = 40 g / m o l S u b s t i t u t i n g t h e v a l u e s i n e q u a t i o n 2 n d . 4.5 = \frac{n_{N a O H}}{1.8 L} n_{N a O H} = 4.5 \times 1.8 L n_{N a O H} = 8.1 m o l e s N o w u \sin g e q u a t i o n 1 s t t o c a l c u l a t e t h e m a s s o f N a O H . 8.1 m o l e s = \frac{m_{N a O H}}{40 g / m o l} m_{N a O H} = (8.1 \times 40) g m_{N a O H} = 324 g \approx 3.2 \times 10^{2} g (answer)$

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